solving a DE with power series: 3y"+xy'-4y=0. Since there is no singular point I can use the regular method (instead of Frobenius). I also have to find the convergence radius. Found recursive relationship, solution and radius will be posted in the first post. Can someone please overview it?

Question

anonymous · Answer

$$DE: 3y"+xy'-4y=0$$

recursive relationship:
$$c _{n+2}= - \frac{ (n-4) cn }{ 3(n+1)(n+2) }$$

what if n=0,1,2,3,...

$$n=0 \rightarrow c _{2} = \frac{ 2 c _{0} }{ 3 }$$
$$n=1 \rightarrow c _{3} = \frac{ c _{1} }{ 3! }$$
$$n=2 \rightarrow c _{4} = \frac{ 2 c _{0} }{ 3^{3} }$$
all even coefficients after this are 0
$$n=3 \rightarrow c _{5} = \frac{ c _{1} }{ 3.5! }$$
$$n=5 \rightarrow c _{7} = \frac{- c _{1} }{ 3^{2}.7! }$$
$$n=7 \rightarrow c _{9} = \frac{3 c _{1} }{ 3^{3}.9! }$$
$$n=9 \rightarrow c _{11} = \frac{-5!! c _{1} }{ 3^{4}.11! }$$

Sorry, haven't worked out the rest yet... confused one ending of one problem I did with this one. Will post what I think is the series next.

anonymous · Answer

There's a mistake for c4. That should be c4=2c0/3^2.3! or c4= c0/3^3

anonymous · Answer

I think the series for the even series with c0 is as follows

$$y _{even}=\frac{2 }{ 3 }c _{0}\sum_{n=0}^{\infty}\frac{ x ^{2n} }{ (2n+1)! }$$

anonymous · Answer

For the uneven series with c1 I have so far

$$y _{uneven}= c _{1} (x+\frac{ x ^{3} }{ 3 }+\frac{ x ^{5} }{ 3. 5! }+\sum_{n=3}^{\infty}\frac{ (-1)^{n} n!!x ^{2n+1}}{ 3^{n-1}(2n+1)!})$$

I'm going to see whether I can rewrite the series with the formulas of double factorials so that I can include the lower part of the series.

anonymous · Answer

That uneven expression was wrong. It should be (n-2)!! in the numenator.

I've come up with this expression. It contains 2 series.

$$y _{uneven}=c _{1}(\sum_{n=0}^{1}\frac{ x ^{n} }{ n! }+\sum_{n=2}^{\infty}\frac{ (-1)^{n}(n-2)!!x ^{2n+1} }{ 3^{n-1} (2n+1)!})$$

I think that the even series summation will not go to infinity but to 2. That makes that 

$$y=\frac{ 2 }{ 3 } c0\sum_{n=0}^{2}\frac{ x ^{2n} }{ (2n+1)! } + c _{1}(\sum_{n=0}^{1}\frac{ x ^{n} }{ n! }+\sum_{n=2}^{\infty}\frac{ (-1)^{n}(n-2)!!x ^{2n+1} }{ 3^{n-1} (2n+1)!})$$

Am I allowed to use finite summation like this?

kinggeorge · Answer

Well, there's nothing wrong with using a finite summation like that. I'll admit, I haven't gone through and checked all of your work, but assuming it's correct, I see no reason why you couldn't use a finite sum.

Of course, it's preferred if you can incorporate it into a single infinite sum, but it shouldn't be required.

anonymous · Answer

thanks, @kinggeorge, I tried to get the large infinite summation of the uneven series to go to n=1 but then it wouldn't result in the action second term. 

For the convergence radius of this I calculated the limit of each series with the ratio test, thinking that I ought to add the resulting limits with each other and have them be smaller than 1.

I found the respective limits:

$$\left| \frac{ x }{ 6 } \right|+\left| \frac{ x }{ 2 } \right| + \left| \frac{ -x }{ 3\sqrt{2}} \right|.0 = \left| \frac{ 4x }{ 6 }  \right|<1$$
$$\frac{ -3 }{ 2 }< x <\frac{ 3 }{ 2 }$$

So R= 3/2

anonymous · Answer

Or should I calculate three times a radius for each limit and take the smallest range allowed for x? 

I just thought that since the series are added up to make y and the convergence of y is asked I needed to take the sum of limits. Anything wrong with that line of thought?

kinggeorge · Answer

Well finite sums will always converge. So you shouldn't need to do any kind of test or those. You only care about the infinite sum when testing convergence.

Oh, and to make it easier on the eyes, instead of writing $$\frac{ 2 }{ 3 } c_0\sum_{n=0}^{2}\frac{ x ^{2n} }{ (2n+1)! }$$it might be better to write$$\frac23c_0(1+\frac{x^2}{3!}+\frac{x^4}{5!}).$$Just a thought.

anonymous · Answer

hmm then I have to make the summation 

$$\sum_{n=0}^{1}$$

because the even series stops after x^2. Everything beyond that is 0.

Well it's not that I have to test for convergence but have to give the radius. The radius for the infinite sum is infinite. Because the limit of the infinite sum = 0. x can be any real number there.

Originally I thought that was the answer, but then I thought, well ok the other 2 certainly converge, but their radius is smaller: there's a limit to what the values of x can be.

kinggeorge · Answer

Sorry for being so slow, dealing with super annoying trolls.

But since finite sums always converge, their radius of convergence of is $\infty$.

anonymous · Answer

Ah, ok... thanks very much! I guess I can close this one then :-)

anonymous · Answer

@KingGeorge based on your post where you suggest to rewrite the even series I have a hypothetical question. Let's say that I might have the following series
$$2c_{1}\sum_{n=0}^{1}\frac{ (n+1)x^{2n+1} }{ 3^{n+1}(2n+1)! }$$

There is an elemntary function hidden there, namely sinh(x). Am I then allowed to rewrite it as 

$$2c_{1} \sinh(x) \sum_{n=0}^{1}\frac{ (n+1) }{ 3^{n+1}}$$